Group Anagrams Using Hash Map in DSA Java - Build from Scratch

Group Anagrams Using Hash Map

📖 Scenario: You work in a bookstore. You have a list of book titles, and you want to group titles that are anagrams of each other. Anagrams are words made by rearranging the letters of another word, like "listen" and "silent".

🎯 Goal: Build a Java program that groups book titles which are anagrams using a hash map.

📋 What You'll Learn

Create a list of book titles with exact values

Create a hash map to group anagrams

Use a loop to process each title and group anagrams

Print the grouped anagrams

💡 Why This Matters

🌍 Real World

Grouping anagrams helps in organizing data like book titles, product names, or words in dictionaries for quick search and analysis.

💼 Career

Understanding hash maps and string manipulation is essential for software developers working on text processing, search engines, and data grouping tasks.

1

Create the list of book titles

Create a List<String> called titles with these exact values: "listen", "silent", "enlist", "hello", "ohlle"

DSA Java

import java.util.*;

public class Main {
    public static void main(String[] args) {
        // Create the list of book titles
        // Your code here
    }
}

Hint

Use Arrays.asList to create the list with the exact titles.

2

Create the hash map to group anagrams

Create a Map<String, List<String>> called anagramGroups and initialize it as a new HashMap<>()

DSA Java

        Map> anagramGroups = new HashMap();
        // Group the titles by their sorted characters

Hint

Use Map<String, List<String>> and new HashMap<>() to create the map.

3

Group the titles by their sorted characters

Use a for loop with variable title to iterate over titles. Inside the loop, create a char[] from title.toCharArray(), sort it using Arrays.sort(), then convert it back to a String called key. Add title to the list in anagramGroups for this key, creating a new list if needed.

DSA Java

        for (String title : titles) {
            char[] chars = title.toCharArray();
            Arrays.sort(chars);
            String key = new String(chars);
            anagramGroups.computeIfAbsent(key, k -> new ArrayList()).add(title);
        }
        // Print the grouped anagrams

Hint

Sort the characters of each title to find the key. Use computeIfAbsent to add titles to the map.

4

Print the grouped anagrams

Use a for loop with variables key and group to iterate over anagramGroups.entrySet(). Inside the loop, print the group list.

DSA Java

Hint

Print each group list on its own line using System.out.println().

Practice

1. What is the main idea behind grouping anagrams using a hash map?

easy

A. Count the length of each word and group by length

B. Group words by their first letter only

C. Use sorted strings as keys to group words that are anagrams

D. Use the original word as the key in the hash map

Solution

1. Step 1: Understand anagram grouping

   Anagrams are words with the same letters in different orders. Sorting letters gives a unique key for each group.

2. Step 2: Use sorted string as hash map key

   By sorting each word and using it as a key, all anagrams map to the same key and can be grouped together.

3. Final Answer:

   Use sorted strings as keys to group words that are anagrams -> Option C

4. Quick Check:

   Sorted string key = group anagrams [OK]

Hint: Sort each word and use as key to group anagrams [OK]

Common Mistakes:

• Grouping by word length only
• Using original word as key without sorting
• Grouping by first letter only

2. Which of the following Java code snippets correctly initializes a hash map to group anagrams?

easy

A. HashMap map = new HashMap<>();

B. HashMap map = new HashMap<>();

C. Map map = new HashMap<>();

D. Map map = new HashMap<>();

Solution

1. Step 1: Identify correct map types for grouping

   We need keys as strings (sorted words) and values as lists of strings (grouped anagrams).

2. Step 2: Check each option's type correctness

   Map map = new HashMap<>(); uses Map, which is correct. Others have wrong key or value types.

3. Final Answer:

   Map map = new HashMap<>(); -> Option D

4. Quick Check:

   Key=String, Value=List<String> for grouping [OK]

Hint: Use Map for grouping anagrams [OK]

Common Mistakes:

• Using int as key type
• Using String as value instead of List<String>
• Using List<String> as key type

3. What is the output of the following Java code snippet?

String[] words = {"eat", "tea", "tan", "ate", "nat", "bat"};
Map map = new HashMap<>();
for (String word : words) {
  char[] chars = word.toCharArray();
  Arrays.sort(chars);
  String key = new String(chars);
  map.computeIfAbsent(key, k -> new ArrayList<>()).add(word);
}
System.out.println(map.values());

medium

A. [[eat, tea, ate], [tan, nat], [bat]]

B. [[eat, tea], [tan, nat], [bat, ate]]

C. [[eat, ate], [tea, tan], [nat, bat]]

D. [[eat], [tea], [tan], [ate], [nat], [bat]]

Solution

1. Step 1: Sort each word to get keys

   "eat", "tea", "ate" all sort to "aet"; "tan", "nat" to "ant"; "bat" to "abt".

2. Step 2: Group words by sorted keys

   Words with key "aet" grouped: [eat, tea, ate]; key "ant": [tan, nat]; key "abt": [bat].

3. Final Answer:

   [[eat, tea, ate], [tan, nat], [bat]] -> Option A

4. Quick Check:

   Sorted keys group anagrams correctly [OK]

Hint: Sort words, group by sorted key, print grouped lists [OK]

Common Mistakes:

• Not sorting words before grouping
• Adding words to wrong groups
• Printing keys instead of grouped values

4. Identify the error in this Java code for grouping anagrams:

String[] words = {"bat", "tab", "eat"};
Map map = new HashMap<>();
for (String word : words) {
  char[] chars = word.toCharArray();
  String key = new String(chars);
  Arrays.sort(chars);
  map.putIfAbsent(key, new ArrayList<>());
  map.get(key).add(word);
}
System.out.println(map.values());

medium

A. Sorting is done after creating the key, so keys are incorrect

B. Using putIfAbsent instead of computeIfAbsent causes error

C. The map should use List<Integer> instead of List<String>

D. The loop should use a while instead of for

Solution

1. Step 1: Check order of sorting and key creation

   Key is created from chars before sorting, so key is unsorted word.

2. Step 2: Identify impact on grouping

   Since keys are unsorted, anagrams get different keys and won't group correctly.

3. Final Answer:

   Sorting is done after creating the key, so keys are incorrect -> Option A

4. Quick Check:

   Sort before key creation to group correctly [OK]

Hint: Sort chars before creating key string [OK]

Common Mistakes:

• Creating key before sorting chars
• Confusing putIfAbsent with computeIfAbsent
• Wrong map value type

5. Given a list of words, how would you modify the grouping anagrams approach to handle case-insensitive grouping (e.g., "Eat" and "tea" are anagrams)?

hard

A. Use the original word as key without any change

B. Convert each word to lowercase before sorting and using as key

C. Group words by their length only

D. Sort words but keep original case for keys

Solution

1. Step 1: Understand case-insensitive grouping

   To treat "Eat" and "tea" as same group, ignore case by converting all words to lowercase.

2. Step 2: Apply lowercase conversion before sorting

   Convert word to lowercase, then sort chars to create key. This ensures case differences don't split groups.

3. Step 3: Group words using lowercase sorted keys

   Use these keys in hash map to group anagrams regardless of original case.

4. Final Answer:

   Convert each word to lowercase before sorting and using as key -> Option B

5. Quick Check:

   Lowercase + sort = case-insensitive grouping [OK]

Hint: Lowercase words before sorting for case-insensitive groups [OK]

Common Mistakes:

• Not converting to lowercase before sorting
• Using original case keys causing split groups
• Grouping by length instead of letters